Is the cohomology of the corresponding Lie algebroid an invariant under equivalence of source-simply-connected Lie groupoids?

Question

Recall the related notions of Lie groupoid, Lie algebroid, generalized morphism of Lie groupoids, and cohomology of Lie algebroid. Henceforth, I will drop the word "Lie" for all those things listed above, because I want to reuse it: there is a functor "Lie" from the 1-category of groupoids to the 1-category of algebroids. There is also a (contravariant) functor "cohomology" from the 1-category of algebroids to the 1-category of graded commutative algebras. However, their composition does not extend to generalized morphisms (which are really 1-morphisms in a 2-category of groupoids).

In particular, the algebroids for equivalent groupoids need not have isomorphic cohomology. A good example is as follows. For any manifold $M$, there is a "pair" groupoid $M\times M \rightrightarrows M$ with object the points in $M$ and a unique morphism between each pair of points. In fact, this groupoid is equivalent to the groupoid $\{\text{pt}\}$ with one object and one morphisms. But $\operatorname{Lie}(M\times M \rightrightarrows M) = {\rm T}M$ is the tangent groupoid, and the cohomology of this algebroid is the de Rham cohomology of $M$, which need not be trivial.

My question is:

If $G_1,G_2$ are two equivalent groupoids, and if both are source-simply-connected, does it follow that the cohomologies of the algebroids $\operatorname{Lie}(G_1)$ and $\operatorname{Lie}(G_2)$ are isomorphic?

There is a converse question, for which I am less optimistic, and that I haven't thought much about myself:

If $G_1,G_2$ are source-simply-connected groupoids and the cohomologies of $\operatorname{Lie}(G_1)$, $\operatorname{Lie}(G_2)$ are isomorphic, does it follow that $G_1,G_2$ are equivalent?

My motivation is the following. There ought to be (but there is not, although almost) a "Lie III theorem" that says that the categories of algebroids and of source-simply-connected groupoids are equivalent. Groupoids present stacks, and the question becomes what "stack-like" thing algebroids present. If the answers to both questions are "yes", then the "stack-like thing" presented by an algebroid just is its cohomology. But probably the answers are not both "yes" — even answers "yes, no" means that, well, the cohomology doesn't entirely determine the stack, but it is an invariant.

Answer 1

I may be misinterpreting what you said (in particular, I don't know what you mean by "source simply connected"), but it sounds like you basically answered your own question in the negative: By employing the pair construction $M \times M \rightrightarrows M$, it suffices to find two simply connected manifolds with nonisomorphic cohomology. The standard examples are the point and the sphere $S^2$.

Answer 2

Here is a counterexample to the second question. Let $\mathfrak g$ be the nonabelian two-dimensional Lie algebra, thought of as an algebroid with one object, and $G$ its simply-connected group, thought of as a groupoid with one object. Then the abelianization of $G$ is the one-dimensional Lie group $\mathbb R$. It is clear that $G$ and $\mathbb R$ are not equivalent as groupoids. Nevertheless, the abelianization map $G\to \mathbb R$ induces an isomorphism on algebroid cohomology: The chain complex computing the cohomology of $\mathfrak g$ is $\mathbb R \overset 0 \to \mathfrak g^* \overset{ [,]^* }\to (\mathfrak g^*)^{\wedge 2}$, and the second arrow is a surjection, hence the cohomology is $\mathbb R, (\mathfrak g / [\mathfrak g,\mathfrak g])^*, 0$. Of course, the cohomology of $\operatorname{Lie}(\mathbb R)$ is $\mathbb R,\mathbb R$, computed by $\mathbb R \overset 0 \to \mathbb R$. The abelianization map $G \to \mathbb R$ induces the injection $\mathbb R \hookrightarrow \mathfrak g^*$ with image the kernel of $[,]^*$.